BetaT

BetaT

Assignment 1 

Profile Assessment

Application 1 naiver strokes flow velocity.

This application calculates the naiver strokes flow velocity.

Naiver Strokes is an equation for Flow Velocity.

Navier–Stokes equations are useful because they describe the physics of many phenomena of scientific and engineering interest. They may be used to model the weather, ocean currents, water flow in a pipe and air flow around a wing. The Navier–Stokes equations in their full and simplified forms help with the design of aircraft and cars, the study of blood flow, the design of power stations, the analysis of pollution, and many other things. Coupled with Maxwell's equations they can be used to model and study magnetohydrodynamics. courtesy of wikipedia ("https://en.wikipedia.org/wiki/Navier%E2%80%93Stokes_equations")

Application Code

 /*
  * Solves 1D convection equation, also known as the one-way wave equation.
  */

1. include <iostream>
2. include <fstream>
3. include <vector>
4. include <iomanip>
5. include <cstdlib>
6. include <chrono>

using namespace std::chrono; using namespace std;

void saveArray(vector< vector<double> > &u, int nx, int nt, double dx, double dt, int c);

void reportTime(const char* msg, steady_clock::duration span) { auto ms = duration_cast<milliseconds>(span); std::cout << msg << " - took - " << ms.count() << " millisecs" << std::endl; }

int main(int argc,char** argv) {

if (argc != 3 )
{

std::cerr << "Invalid number of arguments " << argv[0] << "\n"; return 1;

}
steady_clock::time_point ts, te; 
ts = steady_clock::now();
int nx = std::atoi(argv[1]);
 int nt = std::atoi(argv[2]);
 double dx = double(2)/(nx-1);
 double dt = 0.01;
 int c = 10;

 std::cout << c*dt / dx << "\n";
 // ionitalize 

 vector< vector<double> > u(nx,vector<double> (nt));
 vector< vector<double> > un(nx,vector<double> (nt));

 // Initial condition:

//ts = steady_clock::now();

 for (int i=0; i <= nx-1; i++)
   {
     if (i*dx >= 0.5 && i*dx <= 1)
   {
     u[i][0] = 2;
   }
     else
   {
     u[i][0] = 1;
   }
   }

// te= steady_clock::now(); // reportTime("Fist for loop", te - ts);

 // Finite-difference loop:

//ts = steady_clock::now(); for (int it=1; it<=nt-1; it++)

   {
     for (int k=0; k<=nx-1; k++)
   {
     un[k][it-1] = u[k][it-1];
   }
     for (int i=1; i<=nx-1; i++)
   {
       u[0][it] = un[1][it-1];
       u[i][it] = un[i][it-1] - c*dt/dx*(un[i][it-1]-un[i-1][it-1]);
   }
   }

saveArray(un, nx, nt, dx, dt, c); te = steady_clock::now(); reportTime("Total Time", te - ts);

system("pause"); return 0; } // Save array to file: void saveArray(vector< vector<double> > &u, int nx, int nt, double dx, double dt, int c) {

 ofstream myfile;
 myfile.open("1d_convection02.dat");
 myfile << "%\t" << "nx = "  << nx  << "\tnt = " << nt <<
   "\tdt = " << dt << "\tc = " << c << endl;

 for (int i=0; i<=nx-1; i++)
   {
     myfile << std::setw(8) << std::setfill(' ') << std::left << i*dx << "\t\t";
     for (int j=0; j<=nt-1; j++)
   {
     myfile << std::setw(8) << std::setfill(' ') << std::left << u[i][j] << "\t\t";
   }
     myfile << endl;
   }

}

problem

The problem with this application comes in the main function trying to calculate the finite-difference

The user inputs 2 values which will be used as a reference for the loop.

 // Finite-difference loop:
 for (int it=1; it<=nt-1; it++)
   {
     for (int k=0; k<=nx-1; k++)
   {
     un[k][it-1] = u[k][it-1];
   }
     for (int i=1; i<=nx-1; i++)
   {
     u[0][it] = un[1][it-1];
     u[i][it] = un[i][it-1] - c*dt/dx*(un[i][it-1]-un[i-1][it-1]);
   }
   }

Tests ran with no optimization on linux

By using the command line argument cat /proc/cpuinfo We can find the CPU specs for the VM we are operating linux through. for this test we have: Dual-Core AMD Opteron cpu MHz at 2792

Naiver Equation

n	Time in Milliseconds
100 x 100	24
500 x 500	352
1000 x 1000	1090
2000 x 2000	3936
5000 x 5000	37799
5000 x 10000	65955
10000 x 10000	118682
12500 x 12500	220198

gprof

it gets a bit messy down there, but basically 89.19% of the program is spent in the main() calculating those for loops shown above. The additional time is spent allocating the memory which might cause some slowdown when transferring it to the GPU across the bus int he future.

But the main thing to take away here is that main() is 89.19% and takes 97 seconds.

Each sample counts as 0.01 seconds.

 %   cumulative   self              self     total           
time   seconds   seconds    calls   s/call   s/call  name    
89.19     97.08    97.08                             main
 4.73    102.22     5.14 1406087506     0.00     0.00  std::vector<std::vector<double, std::allocator<double> >, std::allocator<std::vector<double, std::allocator<double> > > >::operator[](unsigned int)
 4.49    107.11     4.88 1406087506     0.00     0.00  std::vector<double, std::allocator<double> >::operator[](unsigned int)

Potential Speed Increase with Amdahls Law

Using Amdahls Law ---- > Sn = 1 / ( 1 - P + P/n )

We can examine how fast out program is capable of increasing its speed.

P = is the part of the program we want to optimize which from above is 89.17% n = the amount of processors we will use. One GPU card has 384 processors or CUDA cores and another GPU we will use has 1020 processor or CUDA cores.

Applying the algorithm gives us.

Amdahls Law for GPU with 384 Cores---- > Sn = 1 / ( 1 - 0.8919 + 0.8919/384 )

                                        Sn = 9.0561125222

Amdahls Law for GPU with 1024 Cores---- > Sn = 1 / ( 1 - 0.8919 + 0.8919/1024 )

                                         Sn = 9.176753777

Therefor According to Amdahls law we can expect a 9x increase in speed.

97 seconds to execute main / 9 amdahls law = 10.7777 seconds to execute after using GPU

Interestingly according to the law the difference in GPU cores does not significantly increase speed. Future tests will confirm or deny these results.

---

Potential Speed Increase with Gustafsons Law

Gustafsons Law S(n) = n - ( 1 - P ) ∙ ( n - 1 )

(Quadro K2000 GPU) S = 380 - ( 1 - .8918 ) * ( 380 - 1 ) = 339.031

(GeForce GTX960 GPU) S = 1024 - ( 1 - .8918 ) * ( 1024 - 1 ) = 913.3114

Using Gustafsons law we see drastic changes in the amount speed increase, this time the additional Cores made a big difference and applying these speed ups we get:

(Quadro K2000 GPU) 97 seconds to execute / 339.031 = 0.29

(GeForce GTX960 GPU) 97 seconds to execute / 913.3114 = 0.11

Tests ran with no optimization on Windows nisghts

System Specifications

Application 2 Calculating Pi

This application is pretty straightforward, it calculates Pi to the decimal point which is given by the user. So an input of 10 vs 100,000 will calculate Pi to either the 10th or 100 thousandth decimal.

problem

Inside the function calculate we have:

void calculate(std::vector<int>& r, int n) { int i, k; int b, d; int c = 0;

for (i = 0; i < n; i++) {

r[i] = 2000;

}

for (k = n; k > 0; k -= 14) {

d = 0; i = k;

for (;;) {

d += r[i] * 10000;

b = 2 * i - 1;

r[i] = d % b;

d /= b;

i--;

if (i == 0) break;

d *= i; }

//printf("%.4d", c + d / 10000); c = d % 10000; } }

I Believe the 2 for loops will cause a delay in the program execution time.

Tests ran with no optimization on linux

for this test the linux VM has: Dual-Core AMD Opteron cpu MHz at 2792

Naiver Equation

n	Time in Milliseconds
1000	2
10000	266
100000	26616
200000	106607
500000	671163

gprof

As with the other application that was profiled it can be a bit hard to read the gprof results. Basically the program spends 87% of the time in the calculate() method and with a problem size of 500,000 it spend a cumulative of 354 seconds. Hopefully we can get this number down.

But the main thing to take away here is that main() is 89.19% and takes 97 seconds. Flat profile:

Each sample counts as 0.01 seconds.

 %   cumulative   self              self     total           
time   seconds   seconds    calls   s/call   s/call  name    
87.39    354.08   354.08        1   354.08   395.84  calculate(std::vector<int, std::allocator<int> >&, int)
10.31    395.84    41.76 678273676     0.00     0.00  std::vector<int, std::allocator<int> >::operator[](unsigned int)

Potential Speed Increase with Amdahls Law

Using Amdahls Law ---- > Sn = 1 / ( 1 - P + P/n )

We can examine how fast out program is capable of increasing its speed.

P = is the part of the program we want to optimize which from above is 87.39% n = the amount of processors we will use. One GPU card has 384 processors or CUDA cores and another GPU we will use has 1020 processor or CUDA cores.

Applying the algorithm gives us.

Amdahls Law for GPU with 384 Cores---- > Sn = 1 / ( 1 - 0.8739 + 0.8739/384 )

                                        Sn = 7.789631

Amdahls Law for GPU with 1024 Cores---- > Sn = 1 / ( 1 - 0.8739 + 0.8739/1024 )

                                         Sn = 7.876904

Therefor According to Amdahls law we can expect a 7.7x to 7.9x increase in speed.

97 seconds to execute main / 7.8 amdahls law = 45.3948 seconds to execute after using GPU

Interestingly the last application had p = 89% (9x speed up) and this application p = 87% (7.8x speed up), 2% made quite a difference.

---

Potential Speed Increase with Gustafsons Law

Gustafsons Law S(n) = n - ( 1 - P ) ∙ ( n - 1 )

(Quadro K2000 GPU) S = 380 - ( 1 - .8739 ) * ( 380 - 1 ) = 332.2081

(GeForce GTX960 GPU) S = 1024 - ( 1 - .8739 ) * ( 1024 - 1 ) = 894.9837

Using Gustafsons law we see drastic changes in the amount speed increase, this time the additional Cores made a big difference and applying these speed ups we get:

(Quadro K2000 GPU) 354 seconds to execute / 332.2081 = 1.065597

(GeForce GTX960 GPU) 354 seconds to execute / 894.9837 = 0.395537

Conclusions with Profile Assessment

Based on the problem we have for both applications which is quadratic(A nested for loop). The time spent processing the main problem which was 89.19% and 87.39%. Plus the amount of time in seconds the program spent on the particular problem which was 97 & 354 seconds. I believe it is feasible to optimize one of these application with CUDA to improve performance.

I will attempt to optimize the naiver strokes flow velocity program as that application is more interesting to me.

Parallelize

Original Problems when converting a program to use the GPU

In order to use the GPU with this program some changes need to be implemented.

The original program was using a two dimensional Vector. unfortunately a vector cannot be copied using *cudaMemCpy* or allocated for using *cudaMalloc*, so there needs to be a change, the vector will be converted into a 2D Array of float** u[][]

Now that the program is iterating trough an Array rather than a Vector the program executes at a much faster speed, but it can still go faster so I will continue optimization with the GPU.

The next problem encountered is like a vector we cannot transfer a 2D array to a *cudaMalloc* or *cudaMemCPy* function call. So the 2D array needs to be converted into a single array. And that brings another problem regarding the original algorithm which is shown below.

for (int i=0; i <= nx-1; i++)

   {
     if (i*dx >= 0.5 && i*dx <= 1)
   {
     u[i][0] = 2;
   }
     else
   {
     u[i][0] = 1;
   }
   }

 // Finite-difference loop:
 for (int it=1; it<=nt-1; it++)
   {
     for (int k=0; k<=nx-1; k++)
   {
     un[k][it-1] = u[k][it-1];
   }
     for (int i=1; i<=nx-1; i++)
   {
     u[0][it] = un[1][it-1];
     u[i][it] = un[i][it-1] - c*dt/dx*(un[i][it-1]-un[i-1][it-1]);
   }

}

In order to get this algorithm to work using a 1D array some changes are made, see below

for (int k = 0; k <= nx - 1; k++)
 if (k*dx >= 0.5 && k*dx <= 1)
 {	
 u[k * nt + 0] = 2;	
 }	
 else	
 {		
 u[k * nt + 0] = 1;	
 }

 for (int it = 1; it <= nx - 1; it++)
 {	
  for (int k = 0; k <= nx - 1; k++)	
  {		
   un[k * nx + it - 1] = u[k * nx + it - 1];	
  }	
  for (int m = 1; m <= nx - 1; m++)	
  {		
   u[0 * nx + it] = un[1 * nx + it - 1];		
   u[m * nx + it] = un[m * nx + it - 1] - c*dt / dx*(un[m * nx + it - 1] - un[(m - 1) * nx + it - 1]);	
  }	

}

It is possible to now iterate through a single dimensional array using the original algorithm.

One more small problem came up, not every element in the array is initialized so there are parts which are pointing to a nullptr garbage in memory. Beause originally the program used a vector, by default all elememnts are initialized to 0, now to solve this problem a separate function is implemented to fill each array index with he value of 0 upon initialization.

After these implementations, testing the code produced the same results as the original program, so it is a positive confirmation that we can proceed to optimizing the cod using the GPU

Parallelizing with 2 Kernels

The kernels have been initialized as a 2D Grid dim3 dGrid(nbx, nbx); AND dim3 dBlock(ntpb, ntpb);

In the first kernel I have Replaced the for loop statement. The goal of this first statement was to set the first value in each column to either 1 or 2 based off the condition in the if statement. The for loop is not needed.

  __global__ void Initalize(float* u, float* un, int nx, int nt, float dx)
     {
       int i = blockIdx.x * blockDim.x + threadIdx.x;
       int j = blockIdx.y * blockDim.y + threadIdx.y;
       
          if (i < nx && j < nx)
              {
                
                if (i*dx >= 0.5 && i*dx <= 1)
                    {
                      u[i * nx] = 2;
                    }
                 else
                     {
                       u[i * nx] = 1;
                     }
              }
      }

This was the tricky part in converting the original code into the kernel. I have removed the 2 inner for loops but kept the outer loop. The program takes 2 arrays. Let us say the X's represent the arrays below

  __global__ void Calculate (float* u, float* un,int nx, int c, float dx, float dt)
    {
                         
     int j = blockIdx.x * blockDim.x + threadIdx.x;
     int i = blockIdx.y * blockDim.y + threadIdx.y;
     // removes from instructions because no need to do this NX amount of times
     float total = c*dt / dx;
                           
         if (i < nx && j < nx)
             {
                    
                for (int it = 1; it <= nx- 1; it++)
                     {
                        if (i != 0 || i < nx )
                              {
                               un[i  * nx + it-1] = u[i  * nx + it-1];
                               __syncthreads();
                               u[it] = un[1 * nx + it - 1];
                               __syncthreads();
                               u[i * nx + it ] = un[i * nx + it- 1] - c*dt / dx* (un[i * nx + it - 1] - un[(i - 1) * nx + it - 1]);
                               __syncthreads();	
                              }	
                      }	
     }

Array 1 Array 2

xxxxx xxxxx

xxxxx xxxxx

xxxxx xxxxx

xxxxx xxxxx

xxxxx xxxxx

 Upon initialization the 1st column of the first array gets set, this will be represented by o's

Array 1 Array 2

oxxxx xxxxx

oxxxx xxxxx

oxxxx xxxxx

oxxxx xxxxx

oxxxx xxxxx

 The next kernel below will execute the following calucations.
 1st:  Array 2 will copy the first column of Array 1

Array 1 Array 2

oxxxx oxxxx

oxxxx oxxxx

oxxxx oxxxx

oxxxx oxxxx

oxxxx oxxxx

 2nd:  Array 1 will set the values in its [0,1] (marked by 2) index to the values in Array 2's [1,0] (marked by  )index.

Array 1 Array 2

o2xxx oxxxx

oxxxx 2xxxx

oxxxx oxxxx

oxxxx oxxxx

oxxxx oxxxx

 3rd:  Next Array 1 will calculate its next column (marked by the 3) by performing a calculation as shown above on Array 2's first column   (marked by 3 ).

Array 1 Array 2

o3xxx 3xxxx

o3xxx 3xxxx

o3xxx 3xxxx

o3xxx 3xxxx

o3xxx 3xxxx

 This process will loop until it has reached end of the array.

CPU VS GPU Loop Comparisons Only

Executing the program again with a problem size of 2000 2000 or 4,000,000 we yield the following results. Keep in mind this is only for the kernel launch or the for-loops executing, not the program as a whole.

Device with compute capability 3.0 found (index 0)

Name: Quadro K2000

Compute Capability: 3.0

Total Global Memory: 2147483648

Max Threads per block: 1024

maxGridSize: 002EF650

maxThreadsDim: 002EF644

Clock Rate in khz: 954000

Fist for loop - took - 0 millisecs
2nd for Loop - took - 0 millisecs
Press any key to continue . . .

As compared to the CPU version of the original program

Initialize arrays loop - took - 17 milliseconds
Fist for loop - took - 1 millisecs
2nd for Loop - took - 15373 millisecs
Press any key to continue . . .

OPTIMIZATION

OVERALL EXECUTION OF PROGRAM FOR CPU, PARALLELIZED GPU AND OPTIMIZED CODE

 TIMES ARE IN MILLISECONDS

 N	        Linux	Visual No Parallel	Parallized	
 2000 ^ 2	1160 	20520	                6749	        	
 5000 ^ 2	28787	127373	                n/a	        	
 10000 ^ 2	124179	522576	                n/a	        	

Windows Display Driver Crash for problem size > 2000 & 2000

When I try to give the program an argument of over 2000 & 2000 it will inform me that the windows dispay driver has crashed and rebooted.

After some research I discovered that this is an issue caused by the kernel taking too long to execute. Windows has a default time limit where it will reset the CUDA GPU if it thinks it is frozen due to the amount of time it is taking to perform its calculations.

This is called the Timeout detection & recovery method (TDR). A potential solution I found on the CUDA programming forum on NVidea's website suggested I try the following in the registry:

 To Change the Graphic device timeout, use the following steps.
  
 Exit all Apps and Programs.
  
 Press the WinKey+R keys to display the Run dialog.
  
 Type  regedit.exe  and click OK to open the registry editor.
  
 Navigate to the following registry key:
  
   HKEY_LOCAL_MACHINE\SYSTEM\CurrentControlSet\Control\GraphicsDrivers
   
    With the GraphicsDrivers key selected, on the Edit menu, click New, 
    and then select the following registry value from the drop-down menu 
    specific to your version of Windows (32 bit, or 64 bit):
  
   
    (NOTE: The TdrDelay name is Case Sensitive)
   
          For 64 bit Windows
           a. Select QWORD (64-bit) value.
           b. Type TdrDelay as the Name and click Enter.
           c. Double-click TdrDelay and add 8 for the Value data and clickOK.
 
 

The above potential solution did not solve my problem.... The second solution I found was to change one of the properties on the GPU device named: kernelExecTimeoutEnabled; This property supposedly controls whether or not the device can be timed out. A value of (1) means it can be timed out, while a value of (0) means it is disabled.

The above also did not solve my issue with the display driver crashing.

Solution to Windows Display Driver Crashing

The best way to prevent this error from happening is to make sure the kernel does not take too long to execute... So I altered my code and switched the Kernel Launch statement from a 2D grid to a 1D grid.

This reduced the number of threads firing in the kernel. In the Calculate Kernel which is below you can see the old one had all the threads from the ( y dimension) sitting idle doing nothing except slowing down the execution.

 __global__ void Calculate (float* u, float* un,int nx, int c, float dx, float dt)
 {
   int j = blockIdx.x * blockDim.x + threadIdx.x;
   int i = blockIdx.y * blockDim.y + threadIdx.y;
   
   
   
         
     if (i < nx && j < nx)
  {
    for (int it = 1; it <= nx- 1; it++)
     {
       if (i != 0 || i < nx )
      {
       un[i  * nx + it-1] = u[i  * nx + it-1];
       __syncthreads();
       u[it] = un[1 * nx + it - 1];
       __syncthreads();
      u[i * nx + it ] = un[i * nx + it- 1] - c * dt/dx * (un[i * nx + it - 1] - un[(i - 1) * nx + it - 1]);
      __syncthreads();
     }
   }	
 }

The code below has been altered to remove the (j) variable and combined the two (if) statements into one, so that we can reduce (Thread Divergence), as well as move the (- c*dt/dx* ) recurring instruction set, and place it into a variable called total, so that each thread is NOT performing the same operation which causes a decrease in performance.

 // kernerl
 __global__ void Calculate(float* u, float* un, int nx, int c, float dx, float dt)
 {
  int i = blockIdx.x * blockDim.x + threadIdx.x;
  
  float total = c*dt / dx;
  
  if (i < nx && i != 0)
  {
   for (int it = 1; it <= nx - 1; it++)
    {
     un[i  * nx + it - 1] = u[i  * nx + it - 1];
     __syncthreads();
     u[it] = un[1 * nx + it - 1];
     __syncthreads();
     u[i * nx + it] = un[i * nx + it - 1] - total * (un[i * nx + it - 1] - un[(i - 1) * nx + it - 1]);
     __syncthreads();
   }
  }
 }

With this optimized code it is now possible to execute with a problem size > 2000 & 2000.

FIRST OPTIMIZATION & Execution Comparison Times

If you have not, please take a look at section 3.1.1.1(just above), as it shows how the first iteration of optimization has been delivered.

Below is a comparison of times from the original CPU to the newly optimized kernel execution. These comaprison times are for the WHOLE execution of the program, not just parts. These include memory transfers, allocation, de-allocation and calculations.

  TIMES ARE IN MILLISECONDS

  N	         Linux	 Visual No Parallel	 Parallized	 Optimized_A	
 (2000 ^ 2)	1160 	| 20520	                | 6749	        | 971	
 (5000 ^ 2)	28787   | 127373	        | n/a	        | 1417	
 (10000 ^ 2)	124179	| 522576	        | n/a	        | 3054

SECOND OPTIMIZATION

The original Initialize Kernel needed some altering. Below is the original:

 __global__ void Initalize(float* u, float* un, int nx, int nt, float dx)
   {
     int i = blockIdx.x * blockDim.x + threadIdx.x;
     int j = blockIdx.y * blockDim.y + threadIdx.y;
              	
       if (i < nx && j < nx)
         {
           if (i*dx >= 0.5 && i*dx <= 1)
              {
                u[i * nx] = 2;
                __syncthreads();
              }
            else
              {
                u[i * nx] = 1;
                __syncthreads();
              }
             }
        }

I removed the variable (j), removed the syncthreads() which were not needed, I also removed the function running on the CPU that initializes all indexes int he arrays to 0, and moved it into the GPU below.

 __global__ void Initalize(float* u, float* un, int nx, int nt, float dx)
 {
   int i = blockIdx.x * blockDim.x + threadIdx.x;
   
     if (i < nx)
       {
         for (int it = 0; it < nx; it++)
           {
             u[i * nx + it] = 0;
             un[i * nx + it] = 0;
           }
          		
              if (i*dx >= 0.5 && i*dx <= 1)
                   u[i * nx] = 2;
              else
                   u[i * nx] = 1;
        }
 }